\(S=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)=\dfrac{a+1}{a}.\dfrac{b+1}{b}\)
\(=\dfrac{a+a+b}{a}.\dfrac{b+a+b}{b}=\dfrac{2a+b}{a}.\dfrac{a+2b}{b}\)
\(=\dfrac{2a^2+4ab+ab+2b^2}{ab}=\dfrac{2\left(a^2+2ab+b^2\right)}{ab}+\dfrac{ab}{ab}\)
\(=\dfrac{2}{ab}+1\)
Ta có \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2+2ab\ge0\Rightarrow2ab\le a^2+b^2\)
\(\Rightarrow4ab\le\left(a+b\right)^2=1\Rightarrow ab\le\dfrac{1}{4}\Rightarrow\dfrac{2}{ab}\ge8\Rightarrow\dfrac{2}{ab}+1\ge9\)
hay S>=9
Dấu = xảy ra khi a=b=1/2
vậy minS=9 khi a=b=1/2
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